Statistics

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Statistics

Hypothesis Tests

Contents Developing Null and Alternative Hypotheses Type I and Type II Errors Population Mean: σ Known Population Mean: σ Unknown Population Proportion Hypothesis Testing and Decision Making Calculating the Probability of Type II Errors Determining the Sample Size for Hypothesis Tests About a Population Mean

Contents Population Proportion Hypothesis Testing and Decision Making Calculating the Probability of Type II Errors Determining the Sample Size for Hypothesis Tests About a Population Mean

STATISTICS in PRACTICE

John Morrell & Company is considered the oldest continuously operating meat manufacturer in the United States. Market research at Morrell provides

management with up to- date information on the company’s various products and how the products compare with competing brands of similar products.

STATISTICS in PRACTICE

One research question concerned whether Morrell’s Convenient Cuisine Beef Pot Roast was the preferred choice of more than 50% of the consumer population. The hypothesis test for the research question is Ho: p ≤ .50 vs. Ha: p > .50

In this chapter we will discuss how to formulate hypotheses and how to conduct tests like the one used by Morrell.

Developing Null and Alternative Hypotheses

Hypothesis testing can be used to determine whether a statement about the value of a population parameter should or should not be rejected.

The null hypothesis, denoted by Ho , is a tentative assumption about a population parameter.


The alternative hypothesis, denoted by Ha, is the opposite of what is stated in the null hypothesis. The alternative hypothesis is what the test is attempting to establish.


Testing Research Hypotheses

Testing the Validity of a Claim

Testing in Decision-Making Situations

Testing Research Hypotheses


The research hypothesis should be expressed as the alternative hypothesis.

The conclusion that the research hypothesis is true comes from sample data that contradict the null hypothesis.


Testing Research Hypotheses Example:

Consider a particular automobile model that currently attains an average fuel efficiency of

24 miles per gallon. A product research group developed a new fuel injection system specifically designed to increase the miles-per-gallon rating.


The appropriate null and alternative hypotheses for the study are: Ho : μ ≤ 24 Ha : μ > 24


Testing Research Hypotheses • In research studies such as these, the null and

alternative hypotheses should be formulated so that the rejection of H0 supports the

research conclusion. The research hypothesis therefore should be expressed as the alternative hypothesis.


Testing the Validity of a Claim• Manufacturers’ claims are usually given the benefit of the doubt and stated as the null hypothesis.• The conclusion that the claim is false comes from sample data that contradict the null hypothesis.


Testing the Validity of a ClaimExample: Consider the situation of a manufacturer of soft drinks who states that it fills two-liter containersof its products with an average of at least 67.6 fluid ounces.


Testing the Validity of a ClaimA sample of two-liter containers will be selected, and the contents will be measured to test the manufacturer’s claim. The null and alternative hypotheses as follows. Ho : μ ≥ 67. 6 Ha : μ < 67. 6


Testing the Validity of a Claim• In any situation that involves testing the validity of a claim, the null hypothesis is generally based on the assumption that the claim is true. The alternative hypothesis is then formulated so that rejection of H0 will provide statistical evidence that the stated assumption is incorrect. Action to correct the claim should be considered whenever H0 is rejected.

Testing in Decision-Making Situations


• A decision maker might have to choose between two courses of action, one associated with the null hypothesis and another associated with the alternative hypothesis.


Testing in Decision-Making SituationsExample: Accepting a shipment of goods from a supplier or returning the shipment of goods to the supplier.Assume that specifications for a particular part require a mean length of two inches per part. The null and alternative hypotheses would be formulated as follows. H0 : μ = 2 Ha : μ ≠ 2


Testing in Decision-Making Situations• If the sample results indicate H0 cannot be

rejected and the shipment will be accepted. • If the sample results indicate H0 should be

rejected the parts do not meet specifications. The quality control inspector will have sufficient evidence to return the shipment to the supplier.


Testing in Decision-Making Situations• We see that for these types of situations,

action is taken both when H0 cannot be rejected and when H0 can be rejected.

Summary of Forms for Null and Alternative Hypotheses about a Population Mean

The equality part of the hypotheses always appears in the null hypothesis.(Follows Minitab)

In general, a hypothesis test about the value of a population mean must take one of the following three forms (where 0 is the hypothesized value of the population mean).

One-tailed(lower-tail)

One-tailed(upper-tail)

Two-tailed

0 0: H

0: aH 0 0: H

0: aH 0 0: H

0: aH

Summary of Forms for Null and Alternative Hypotheses about a Population Mean

Example: Metro EMS

Null and Alternative Hypotheses

Operating in a multiplehospital system with approximately 20 mobile medicalunits, the service goal is to respond to medicalemergencies with a mean time of 12 minutes or less.

A major west coast city providesone of the most comprehensiveemergency medical services inthe world.

The director of medical serviceswants to formulate a hypothesistest that could use a sample ofemergency response times todetermine whether or not theservice goal of 12 minutes or lessis being achieved.

Example: Metro EMS

Null and Alternative Hypotheses

Null and Alternative HypothesesThe emergency service is meetingthe response goal; no follow-upaction is necessary.The emergency service is notmeeting the response goal;appropriate follow-up action isnecessary.

H0:

Ha:

where: μ = mean response time for the population of medical emergency requests

Type I and Type II Errors

Because hypothesis tests are based on sample data, we must allow for the possibility of errors. A Type I error is rejecting H0 when it is true. Applications of hypothesis testing that only control the Type I error are often called significance tests.


Type I error The probability of making a Type I error

when the null hypothesis is true as an equality is called the level of significance.

The Greek symbol α(alpha) is used to denote the level of significance, and common choices for α are .05 and .01.


Type I error If the cost of making a Type I error is high,

small values of α are preferred. If the cost of making a Type I error is not too high, larger values of α are typically used.


A Type II error is accepting H0 when it is false. It is difficult to control for the probability of making a Type II error. Most applications of hypothesis testing control for the probability of making a Type I error. Statisticians avoid the risk of making a Type II error by using “do not reject H0” and not “accept H0”.

Type II error


CorrectDecision Type II Error

CorrectDecisionType I ErrorReject H0

(Conclude > 12)

Not reject H0(Conclude < 12)

H0 True( < 12)

H0 False( > 12)Conclusion

Population Condition

Steps of Hypothesis Testing

Step 1. Develop the null and alternative hypotheses.Step 2. Specify the level of significance α .Step 3. Collect the sample data and compute the test statistic.

p-Value Approach

Step 4. Use the value of the test statistic to compute the p-value.

Step 5. Reject H0 if p-value < α.

Critical Value ApproachStep 4. Use the level of significance to

determine the critical value and the rejection rule.

Step 5. Use the value of the test statistic and the rejection rule to determine whether to

reject H0.

Steps of Hypothesis Testing

p-Value Approach toOne-Tailed Hypothesis Testing

Reject H0 if the p-value < α .

The p-value is the probability, computed using the test statistic, that measures the support (or lack of support) provided by the sample for the null hypothesis. The p-value is also called the observed level of significance. If the p-value is less than or equal to the level of significance α , the value of the test statistic is in the rejection region.

Critical Value Approach to One-Tailed Hypothesis Testing

The test statistic z has a standard normal probability distribution. We can use the standard normal probability distribution table to find the z-value with an area of α in the lower (or upper) tail of the distribution.

Critical Value Approach to One-Tailed Hypothesis Testing

The value of the test statistic that established the boundary of the rejection region is called the critical value for the test. The rejection rule is:• Lower tail: Reject H0 if z < - z α • Upper tail: Reject H0 if z > z α

One-Tailed Hypothesis Testing Population Mean: σ Known One-tailed tests about a population mean

take one of the following two forms. Lower Tail Test Upper Tail Test

0

00

::

aHH

0

00

::

aHH

p-Value Approach

p-value 7

0 -za = -1.28

a = .10

z z =-1.46

Lower-Tailed Test About a Population Mean:s Known

Samplingdistribution of z x

n

0/

p-Value < a ,so reject H0.

p-Value Approach

p-Value

0 za = 1.75

a = .04

z z =2.29


n

0/

p-Value < a ,so reject H0.

Lower-Tailed Test About a Population Mean : s Known

a

0 za = 1.28

Reject H0

Do Not Reject H0

z


n

0/

Critical Value Approach

Lower-Tailed Test About a Population Mean : s Known

a

0 za = 1.645

Reject H0

Do Not Reject H0

z


n

0/

Upper-Tailed Test About a Population Mean : s Known


One-Tailed Hypothesis Testing

Example: The label on a large can of Hilltop Coffee

states that the can contains 3 pounds of coffee. The Federal Trade Commission (FTC) knows that Hilltop’s production process cannot place exactly 3 pounds of coffee in each can, even if the mean filling weight for the population of all cans filled is 3 pounds per can.

One-Tailed Hypothesis TestingExample: However, as long as the population mean

filling weight is at least 3 pounds per can, the rights of consumers will be protected.

One-Tailed Hypothesis TestingStep 1. Develop the null and alternative hypotheses. denoting the population mean filling weight and the hypothesized value of the population mean is μ0 = 3. The null and alternative hypotheses are H0 : μ ≥ 3 Ha : μ < 3 Step 2. Specify the level of significance α. we set the level of significance for the hypothesis test at α= .01

One-Tailed Hypothesis TestingStep 3. Collect the sample data and compute

the test statistic. Test statistic:

Sample data: σ = .18 and sample size n=36, = 2.92 pounds.

nx

z o

/

x

One-Tailed Hypothesis TestingStep 3. Collect the sample data and compute

the test statistic.

We have z = -2.67

03.3

xx

zx

o

One-Tailed Hypothesis Testing p-Value ApproachStep 4. Use the value of the test statistic to

compute the p-value. Using the standard normal distribution table, the area between the mean and z = - 2.67 is .4962. The p-value is .5000 - .4962 = .0038,

Step 5. p-value = .0038 < α = .01 Reject H0.

One-Tailed Hypothesis Testing p-value for The Hilltop Coffee Study when sample mean = 2.92 and z = -2 .67.x

One-Tailed Hypothesis TestingCritical Value ApproachStep 4. Use the level of significance to determine

the critical value and the rejection rule.The critical value is z = -2.33.

One-Tailed Hypothesis Testing

Step 5. Use the value of the test statistic and the rejection rule to determine whether to reject H0.

We will reject H0 if z < - 2.33. Because z = - 2.67 < 2.33, we can reject H0

and conclude that Hilltop Coffee is under filling cans.

Example: Metro EMS The response times for a randomsample of 40 medical emergencieswere tabulated. The sample meanis 13.25 minutes. The populationstandard deviation is believed tobe 3.2 minutes.

One-Tailed Tests About a Population Mean : s Known

Example: Metro EMS The EMS director wants toperform a hypothesis test, with a.05 level of significance, to determinewhether the service goal of 12 minutes or less is being achieved.

One-Tailed Tests About a Population Mean : s Known

3. Compute the value of the test statistic.2. Specify the level of significance.

1. Develop the hypotheses.

a = .05

H0: Ha:

p -Value and Critical Value Approaches

One-Tailed Tests About a Population Mean: s Known

13.25 12 2.47/ 3.2/ 40xz

n

4. Compute the p –value.

5. Determine whether to reject H0.

We are at least 95% confident that Metro EMS is not meeting the response goal of 12 minutes.

p –Value Approach

For z = 2.47, cumulative probability = .9932.p–value = 1 .9932 = .0068

Because p–value = .0068 < α = .05, we reject H0.


p-value

0 za =1.645

a = .05

z z =2.47


n

0/

p –Value Approach


4. Determine the critical value and rejection rule.


We are at least 95% confident that Metro EMS is not meeting the response goal of 12 minutes.

Because 2.47 > 1.645, we reject H0.


For α = .05, z.05 = 1.645Reject H0 if z > 1.645


Two-Tailed Hypothesis Testing In hypothesis testing, the general form for a

two-tailed test about a population mean is as follows:

0

00

::

aHH

p-Value Approach toTwo-Tailed Hypothesis Testing

Compute the p-value using the following three steps:

2. If z is in the upper tail (z > 0), find the area under the standard normal curve to the right of z. If z is in the lower tail (z < 0), find the area under the standard normal curve to the left of z.

1. Compute the value of the test statistic z.

p-Value Approach toTwo-Tailed Hypothesis Testing

The rejection rule: Reject H0 if the p-value < a .

3. Double the tail area obtained in step 2 to obtain the p –value.

Critical Value Approach to Two-Tailed Hypothesis Testing

The critical values will occur in both the lower and upper tails of the standard normal curve.

The rejection rule is: Reject H0 if z < -za/2 or z > za/2

Use the standard normal probability distribution table to find za/2 (the z-value with an area of a/2 in the upper tail of the distribution).

Two-Tailed Hypothesis Testing Example: The U.S. Golf Association

(USGA) establishes rules that manufacturers of golf equipment must meet if their products are to be acceptable for use in USGA events.

MaxFlight produce golf balls with an average distance of 295 yards.

Two-Tailed Hypothesis Testing When the average distance falls below 295

yards, the company worries about losing sales because the golf balls do not provide as much distance as advertised.

Two-Tailed Hypothesis Testing When the average distance passes 295 yards,

MaxFlight’s golf balls may be rejected by the USGA for exceeding the overall distance standard concerning carry and roll. A hypothesized value of μ0= 295 and the null

and alternative hypotheses test are as follows: H0 : μ = 295 VS Ha : μ≠ 295

Two-Tailed Hypothesis Testing If the sample mean is significantly less

than 295 yards or significantly greater than 295

yards, we will reject H0.

x

Two-Tailed Hypothesis TestingStep 1. Develop the null and alternative hypotheses. A hypothesized value of μ0 = 295 and

H0 : μ = 295 Ha : μ≠ 295Step 2. Specify the level of significance α. we set the level of significance for the

hypothesis test at α = .05

Two-Tailed Hypothesis TestingStep 3. Collect the sample data and compute the

test statistic. Test statistic:

Sample data: σ = 12 and sample size n=50, = 297.6 yards. We have

nxz o

/

53.150/122956.297

/

nxz o

x

Two-Tailed Hypothesis Testing p-Value ApproachStep 4. Use the value of the test statistic to

compute the p-value. The two-tailed p-value in this case is given by P(z < -1.53) + P(z > 1.53)= 2(.0630) = .1260.Step 5. We do not reject H0 because the p-value = .1260 > .05. No action will be taken to adjust manufacturing process.

Two-Tailed Hypothesis Testing p-Value for The Maxflight Hypothesis Test

Two-Tailed Hypothesis Testing

Critical Value ApproachStep 4. Use the level of significance to determine

the critical value and the rejection rule. The critical values are - z.025 = -1.96 and z.025 = 1.96.

Two-Tailed Hypothesis Testing

Step 5. Use the value of the test statistic and the rejection rule to determine whether to reject H0.

We will reject H0 if z < - 1.96 or if z > 1.96 Because the value of the test statistic is

z=1.53, the statistical evidence will not permit us to reject the null hypothesis

at the .05 level of significance.

Example: Glow Toothpaste Two-Tailed Test About a Population Mean: s

Known

oz.Glow

The production line for Glow toothpasteis designed to fill tubes with a mean weightof 6 oz. Periodically, a sample of 30 tubeswill be selected in order to check thefilling process.

Example: Glow Toothpaste Two-Tailed Test About a Population Mean: s

Known oz.Glow

Quality assurance procedures call forthe continuation of the filling process if thesample results are consistent with the assumptionthat the mean filling weight for the population of toothpastetubes is 6 oz.; otherwise the process will be adjusted.

Example: Glow Toothpaste Two-Tailed Test About a Population Mean:

s Known

oz.Glow

Perform a hypothesis test, at the .03level of significance, to help determinewhether the filling process should continueoperating or be stopped and corrected.

Assume that a sample of 30 toothpaste tubes provides a sample mean of 6.1 oz. The population standard deviation is believed to be 0.2 oz.

1. Determine the hypotheses. 6:6:0

aHH

2. Specify the level of significance.

3. Compute the value of the test statistic.

α = .03

p –Value and Critical Value Approaches

Glow

Two-Tailed Tests About a Population Mean:s Known

0 6.1 6 2.74/ .2/ 30xz

n

Glow

Two-Tailed Tests About a Population Mean : s Known


p –Value Approach4. Compute the p –value.

For z = 2.74, cumulative probability = .9969p–value = 2(1 - .9969) = .0062

Because p–value = .0062 < α = .03, we reject H0. We are at least 97% confident that the mean filling weight of the toothpaste tubes is not 6 oz.

GlowTwo-Tailed Tests About a

Population Mean : s Known

a/2 = .015

0za/2 = 2.17

z

a/2 = .015

p-Value Approach

-za/2 = -2.17z = 2.74z = -2.74

1/2p -value= .0031

1/2p -value= .0031


Glow


We are at least 97% confident that the mean filling weight of the toothpaste tubes is not 6 oz.


For a/2 = .03/2 = .015, z.015 = 2.174. Determine the critical value and rejection rule.

Reject H0 if z < -2.17 or z > 2.17

Two-Tailed Tests About a Population Mean : s Known

a/2 = .015

0 2.17

Reject H0Do Not Reject H0

z

Reject H0

-2.17

Glow



n

0/

Two-Tailed Tests About a Population Mean : σ Known

a/2 = .015

Summary of Hypothesis Tests about a Population Mean: σ Known Case

Confidence Interval Approach toTwo-Tailed Tests About a Population Meant

Select a simple random sample from the population and use the value of the sample mean to develop the confidence interval for the population mean . (Confidence intervals are covered in Chapter 8.) If the confidence interval contains the hypothesized value 0 do not reject H0. Otherwise, reject H0.

xx

The 97% confidence interval for is

/ 2 6.1 2.17(.2 30) 6.1 .07924x zna

Confidence Interval Approach toTwo-Tailed Tests About a Population Mean Glo

w

Because the hypothesized value for thepopulation mean, 0 = 6, is not in this interval,the hypothesis-testing conclusion is that thenull hypothesis, H0: = 6, can be rejected.

or 6.02076 to 6.17924

Test Statistic

Tests About a Population Mean: Unknown

t xs n

0

/

This test statistic has a t distribution with n - 1 degrees of freedom.

Rejection Rule: p -Value Approach

H0: Reject H0 if t ta

Reject H0 if t ta

Reject H0 if t < - ta or t > ta

H0:

H0:


Rejection Rule: Critical Value ApproachReject H0 if p –value < a

p -Values and the t Distribution The format of the t distribution table provided in most statistics textbooks does not have sufficient detail to determine the exact p-value p-value for a hypothesis test. However, we can still use the t distribution table to identify a range for the p-value. An advantage of computer software packages is that the computer output will provide the p-value for the t distribution.

Tests About a Population Mean: Unknown Example: A business travel magazine wants

to classify transatlantic gateway airports according to the mean rating for the population of business travelers.

A rating scale with a low score of 0 and a high score of 10 will be used, and airports with a population mean rating greater than 7 will be designated as superior service airports.


The null and alternative hypotheses for this upper tail test are as follows:

H0 : μ ≤ 7 Ha : μ > 7


Test Statistic

The sampling distribution of t has n – 1 df. We have =7.25, s = 1.052, and n = 60, and

the value of the test statistic is x

t xs n

0

/

p-Value Approach Using Table 2 in Appendix B, the t

distribution with 59 degrees of freedom.

We see that t 1.84 is between 1.671 and 2.001. Although the table does not provide the exact p-value, the values in the “Area in Upper Tail” row show that the p-value must be less than .05 and greater than .025.

MINITAB OUTPUT

The test statistic t = 1.84, and the exact p-value is .035 for the Heathrow rating

hypothesis test. A p-value = .035 < .05 leads to the rejection

of the null hypothesis.


The rejection rule is thus Reject H0 if t > 1.671 With the test statistic t = 1.84 > 1.671, H0

is rejected.

A State Highway Patrol periodically samplesvehicle speeds at various locationson a particular roadway. The sample of vehicle speedsis used to test the hypothesis

Example: Highway Patrol One-Tailed Test About a Population Mean:

Unknown

The locations where H0 is rejected are deemedthe best locations for radar traps.

H0: m < 65

At Location F, a sample of 64 vehicles shows amean speed of 66.2 mph with astandard deviation of4.2 mph. Use a = .05 totest the hypothesis.

Example: Highway Patrol One-Tailed Test About a Population Mean: s

Unknown

One-Tailed Test About a Population Mean : Unknown

1. Determine the hypotheses.



a = .05


H0: < 65Ha: > 65

0 66.2 65 2.286/ 4.2/ 64

xts n

One-Tailed Test About a Population Mean : Unknown

p –Value Approach

4. Compute the p –value.

For t = 2.286, the p–value must be less than .025 (for t = 1.998) and greater than .01 (for t = 2.387). .01 < p–value < .025

One-Tailed Test About a Population Mean: Unknown

p –Value Approach


Because p–value < a = .05, we reject H0.

We are at least 95% confident that the mean speed of vehicles at Location F is greater than 65 mph.


For a = .05 and d.f. = 64 – 1 = 63, t.05 = 1.669

4. Determine the critical value and rejection rule.

Reject H0 if t > 1.669


Critical Value Approach5. Determine whether to reject H0.

We are at least 95% confident that the mean speed of vehicles at Location F is greater than 65 mph. Location F is a good candidate for a radar trap.



a

0 ta =1.669

Reject H0

Do Not Reject H0

t



Summary of Hypothesis Tests about a Population Mean: Unknown Case

The equality part of the hypotheses always appears in the null hypothesis. In general, a hypothesis test about the value of a population proportion p must take one of the following three forms (where p0 is the hypothesized value of the population proportion).

A Summary of Forms for Null and Alternative Hypotheses About a Population Proportion

A Summary of Forms for Null and Alternative Hypotheses About a Population Proportion

One-tailed(lower tail)

One-tailed(upper tail)

Two-tailed

0 0: H p p

0: aH p p0: aH p p0 0: H p p 0 0: H p p

0: aH p p

Test Statisticz p p

p

0

pp p

n 0 01( )

Tests About a Population Proportion

where:

assuming np > 5 and n(1 – p) > 5

Rejection Rule: p –Value Approach

H: p p Reject H0 if z > z

Reject H0 if z < -z

Reject H0 if z za or z z

H: p p

H0: p p


Reject H0 if p –value < α Rejection Rule: Critical Value Approach

Upper Tail Test About aPopulation Proportion

Example: Pine Creek golf course Over the past year, 20% of the players at

Pine Creek were women. A special promotion designed to attract

women golfers. One month after the promotion was implemented.

The null and alternative hypotheses are as follows: H0: p ≤ .20 Ha: p > .20


level of significance of α = .05 be used. Test statistics

npp

ppz)1( 00

0


level of significance of α = .05 be used. Suppose a random sample of 400 players was selected, and that 100 of the players were

women. The proportion of women golfers in the sample is and the value of the test statistic is

25.400100

p

50.202.05.

400)20.1(20.

20.25.

z


p-Value Approach the p-value is the probability that z is greater

than or equal to z = 2.50. Thus, the p-value for the test is .5000 – .4938 = .0062.


A p-value = .0062 < .05 gives sufficient statistical evidence to reject H0 at the .05 level of significance.

Critical Value Approach The critical value corresponding to an area

of .05 in the upper tail of a standard normal distribution is z.05 = 1.645. To reject H0 if

z ≥ 1.645. Because z = 2.50 > 1.645, H0 is rejected.

For a Christmas and New Year’s week, theNational Safety Council estimated that500 people would be killed and 25,000injured on the nation’s roads. TheNSC claimed that 50% of theaccidents would be caused bydrunk driving.

Example: National Safety Council

Two-Tailed Test About aPopulation Proportion

A sample of 120 accidents showed that67 were caused by drunk driving. Usethese data to test the NSC’s claim withα = .05.

Example: National Safety Council


1. Determine the hypotheses.


a = .05


0: .5H p

: .5aH p


a commonerror is using in this formula

p



0 (67/ 120) .5 1.28.045644p

p pz

0 0(1 ) .5(1 .5) .045644120pp p

n


p -Value Approach

4. Compute the p -value.

5. Determine whether to reject H0.Because p–value = .2006 > a = .05, we cannot reject H0.

For z = 1.28, cumulative probability = .8997p–value = 2(1 - .8997) = .2006





For a/2 = .05/2 = .025, z.025 = 1.964. Determine the critical value and rejection rule.

Reject H0 if z < -1.96 or z > 1.96

Because 1.278 > -1.96 and < 1.96, we cannot reject H0.


Summary of Hypothesis Tests about a Population Proportion

Hypothesis Testing and Decision Making

In many decision-making situations the decision maker may want, and in some cases may be forced, to take action with both the conclusion do not reject H0 and the conclusion

reject H0. In such situations, it is recommended that the hypothesis-testing procedure be extended to include consideration of making a Type II error.

Hypothesis Testing and Decision Making Example: A quality control manager must decide to

accept a shipment of batteries from a supplier or to return the shipment because of poor quality.

Suppose the null and alternative hypotheses about the population mean follow.

H0 : μ ≥ 120 Ha : μ < 120

Hypothesis Testing and Decision Making

If H0 is rejected, the appropriate action is to return the shipment to the supplier.

If H0 is not rejected, the decision maker must still determine what action should be taken. In such decision-making situations, it is

recommended to control the probability of making a Type II error. Because knowledge

of the probability of making a Type II error will be helpful.

Calculating the Probability of a Type II Error in Hypothesis Tests About a Population Mean

1. Formulate the null and alternative hypotheses.

3. Using the rejection rule, solve for the value of the sample mean corresponding to the critical value of the test statistic.

2. Using the critical value approach, use the level of significance to determine the critical value and the rejection rule for the test.

5. Using the sampling distribution of for a value of μ satisfying the alternative hypothesis, and the acceptance region from step 4, compute the

probability that the sample mean will be in the acceptance region. (This is the probability of making a Type II error at the chosen

level of μ.)

x


4. Use the results from step 3 to state the values of the sample mean that lead to the acceptance of H0; this defines the acceptance region.


Example: Batteries’ Quality Suppose the null and alternative hypotheses

are H0 : μ ≥ 120 Ha : μ < 120 level of significance of α = .05. Sample size n=36 and σ = 12 hours.


The critical value approach and z.05 = 1.645, the rejection rule is

Reject H0 if z ≤ -1.645


The rejection rule indicates that we will reject H0 if

That indicates that we will reject H0 if

and accept the shipment whenever >

116.71.x

645.136/12

120

xz

71.11636

12645.1120

x


Compute probabilities associated with making a Type II error (whenever the true mean is less than 120 hours and we make the decision to accept H0: μ ≥ 120.)


If μ =112 is true, the probability of making a Type II error is the probability that the sample mean is greater than 116.71 when μ = 112, that is, P( ≥ 116.71 | μ = 112) =?

The probability of making a Type II error (β ) when μ = 112 is .0091 .

xx

Calculating the Probability of a Type II Error in Hypothesis Tests About a Population Mean We can repeat these calculations for other

values of μ less than 120.


The power of the test = the probability of correctly rejecting H0 when it is false. For any particular value of μ, the power is 1-β.

Power Curve for The Lot-Acceptance Hypothesis Test

Example: Metro EMS (revisited) Recall that the response times fora random sample of 40 medicalemergencies were tabulated. Thesample mean is 13.25 minutes.The population standard deviationis believed to be 3.2 minutes.

Calculating the Probability of a Type II Error

Example: Metro EMS (revisited) The EMS director wants to

perform a hypothesis test, with a.05 level of significance, to determinewhether or not the service goal of 12 minutes or less is being achieved.


12 1.6453.2/ 40xz

3.212 1.645 12.832340x

3. Value of the sample mean that identifies the rejection region:

2. Rejection rule is: Reject H0 if z > 1.6451. Hypotheses are: H0: μ and Ha:μ


4. We will accept H0 when < 12.8323x

12.0001 1.645 .9500 .050012.4 0.85 .8023 .197712.8 0.06 .5239 .476112.8323 0.00 .5000 .500013.2 -0.73 .2327 .767313.6 -1.52 .0643 .935714.0 -2.31 .0104 .9896

12.83233.2/ 40z

Values of b 1-b

5. Probabilities that the sample mean will be in the acceptance region:




When the true population mean m is close to the null hypothesis value of 12, there is a high probability that we will make a Type II error.

Observations about the preceding table:

Example: = 12.0001, b = .9500

Example: = 14.0, b = .0104



When the true population mean m is far above the null hypothesis value of 12, there is a low probability that we will make a Type II error.

Power of the Test The probability of correctly rejecting H0 when it is false is called the power of the

test. For any particular value of μ, the power is 1 – β. We can show graphically the power associated with each value of μ; such a graph is called a power curve. (See next slide.)

Power Curve

0.000.100.200.300.400.500.600.700.800.901.00

11.5 12.0 12.5 13.0 13.5 14.0 14.5

Prob

abili

ty o

f Cor

rect

lyRe

ject

ing

Nul

l Hyp

othe

sis

H0 False

The specified level of significance determines the probability of making a Type I error.

By controlling the sample size, the probability of making a Type II error is controlled.

Determining the Sample Size for a Hypothesis Test About a Population Mean


Example: how a sample size can be determined for the lower tail test about a population mean.

H0 : μ ≥ μ0

Ha : μ < μ0

Let α be the probability of a Type I error and zα and zβ are the z value corresponding to an area of α and β, respectively in the upper tail of the standard normal distribution.


Determining the Sample Size for a Hypothesis Test About a Population Mean we compute c using the following formulas

nzc a 0

nzc b 0

nz

nz a

ba 0


To determine the required sample size, we solve for the n as follows.

nz

nza

ba 0

n

zza

ba )(

0

)()(

0 a

zzn

ba

a

a

x

x

SamplingDistribution of whenH0 is trueand m = m0

x

Samplingdistribution of whenH0 is falseand a > 0

x

Reject H0

b

c

c

H0:

Ha:

x n


whereza = z value providing an area of a in the

tailzb = z value providing an area of b in the

tail = population standard deviation0 = value of the population mean in H0

a = value of the population mean used for the Type II error

nz z

a

( )( )a b

2 2

02n

z z

a

( )( )a b

2 2

02


Note: In a two-tailed hypothesis test, use za /2 not za

Relationship Among a, b, and n Once two of the three values are known, the

other can be computed. For a given level of significance a,

increasing the sample size n will reduce b. For a given sample size n, decreasing a will

increase b, whereas increasing a will decrease b


Let’s assume that the director of medical services makes the following statements

about the allowable probabilities for the Type I and Type II errors:

• If the mean response time is = 12 minutes, I am willing to risk an a = .05 probability of rejecting H0.• If the mean response time is 0.75 minutes over

the specification ( = 12.75), I am willing to risk a b = .10 probability of not rejecting H0.


Givena = .05, b = .10za = 1.645, zb = 1.280 = 12, a = 12.75 = 3.2

2 2 2 2

2 20

( ) (1.645 1.28) (3.2) 155.75 156( ) (12 12.75)a

z zn a b

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